JEE Advanced | 2013ThermochemistryHard

Question

The standard enthalpies of formation of CO2(g), H2O(l) and glucose(s) at 250C are -400 kJ/mol, -300 kJ/mol & -1300 kJ/mol respectively. The standard enthalpy of combustion per gram of glucose at 250C is -

Options

A.+ 2900 kJ
B.- 2900 kJ
C.- 16.11 kJ
D.+ 16.11 kJ

Solution

Combustion reaction for glucose is
C6H12O6(s) + 6O2(g)→ 6CO2(g) + 6H2O(l)
ᐃHr = 6(-400) + 6(-300) - (-1300)
= -2900 KJ/mol KJ / gm
= -16.11 KJ/gm

Create a free account to view solution

View Solution Free
Topic: Thermochemistry·Practice all Thermochemistry questions

More Thermochemistry Questions

(i) Cis - 2 - butene → trans - 2 - butene, ᐃH1(ii) Cis - 2 - butene → 1 - butene, ᐃH2(iii) Trans...The ᐃHfo for CO2(g), CO(g) and H2O(g) are - 395.5, - 110.5 and - 241.8 kJ mol-1 respectively. The standard enthalp...Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below: The energy involved...Which of the following statement is incorrect about internal energy ?...What will be the heat of formation of methane, if the heat of combustion of carbon is′-x′ kJ, heat of format...